∆ABC is such that AB = 3 cm, BC = 2 cm and CA = 2.5 cm. If ∆DEF ∼ ∆ABC and EF = 4 cm, then perimeter of ∆DEF is

Given: In $\triangle A B C, A B=3 \mathrm{~cm}, B C=2 \mathrm{~cm}, C A=2.5 \mathrm{~cm} . \triangle D E F \sim \triangle A B C$ and $E F=4 \mathrm{~cm} .$

To find: Perimeter of ΔDEF.

We know that if two triangles are similar, then their sides are proportional

Since ΔABC and ΔDEF are similar,

$\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{CA}}{\mathrm{FD}}$

$\frac{3}{\mathrm{DE}}=\frac{2}{4}=\frac{2.5}{\mathrm{FD}}$

$\frac{3}{\mathrm{DE}}=\frac{2}{4}$

$\mathrm{DE}=6 \mathrm{~cm}$.....(1)

$\frac{2}{4}=\frac{2.5}{\text { FD }}$

$\mathrm{FD}=5 \mathrm{~cm}$.....(2)

From (1) and (2), we get

Perimeter of ΔDEF =  DE + EF + FD = 6 + 4 +5 = 15 cm

Hence the correct answer is (b)